Ask question

Ask question

All categories

2 years ago

9

A celebrated Mark Twain story has motivated contestants in the Calaveras County Jumping Frog Jubilee, where frog jumps as long a

s 2.20 m have been recorded. If a frog jumps 2.20 m and the launch angle is 36.5°, find the frog's launch speed and the time the frog spends in the air. Ignore air resistance.
(a)the frog's launch speed (in m/s)

(b)the time the frog spends in the air (in s)

1 answer:

IrinaK [193]2 years ago

4 0

The frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.

To find the answer, we need to know about the time of flight and range of projectile motion.

<h3>What's the expression of range of a projectile motion?</h3>

Range = U²× sin(2θ)/g
U= initial velocity, θ= angle of projectile and g= acceleration due to gravity
U=√{Range×g/sin(2θ)}
Here, range= 2.20m, = 36.5°
U= √{2.20×9.8/sin(73)}

U= √{2.20×9.8/sin(73)} = 22.5m/s

<h3>What's the expression of time of flight in projectile motion?</h3>

Time of flight= (2×U×sinθ)/g
So, T= (2×22.5×sin36.5°)/9.8

= 2.73 s

Thus, we can conclude that the frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.

Learn more about the range and time period of projectile motion here:

brainly.com/question/24136952

#SPJ1

You might be interested in

A load of bricks is being lifted by a crane at a steady velocity of 5.0m/s when one bricks falls off 6.0 m above the ground

Ganezh [65]

LOOK OUT BELOW ! ! !

3 0

3 years ago

A liquid is poured into a vessel to a depth of 16cm when viewed from the top, the bottom appears to be raised 4cm. What is the r

zalisa [80]

Answer:

Solution

Verified by Toppr

Correct option is

C

3 cm

RI=apparent depthreal depth

Substituting, 34=apparentdepth12

Therefore, apparent depth=412×3=9

The height by which it appears to be raised is 12−9=3cm

Was this answer helpful?

71

0

SIMILAR QUESTIONS

A coin is placed at the bottom of a glass tumbler and then water is added. It appeared that the depth of the coin has been reduced because

Medium

View solution

>

A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again?

3 0

3 years ago

suppose that 273 g of one of the substances listed above displaces 26 mL of water. What is the substance?

guajiro [1.7K]

<span>The unknown substance is silver. I don't see a list of available substances, but let's see if there's something reasonable available that will match. First, let's calculate the density of the unknown substance. Density is mass per volume, so 273 g / 26 mL = 10.5 g/mL Looking up a list of elements sorted by density, I see the following: 10.07 Actinium 10.22 Molybdenum 10.5 Silver 11.35 Lead And silver at 10.5 g/ml is a very nice match for the unknown substances' density of 10.5 g/ml.</span>

6 0

4 years ago

Read 2 more answers

A proton, starting from rest, accelerates through a potential difference of 1.0 kV and then moves into a magnetic field of 0.040

Minchanka [31]

Answer:

r = 0.11 m

Explanation:

The radius of the proton's resulting orbit can be calculated equaling the force centripetal (Fc) with the Lorentz force ( $F_{B}$ ), as follows:

$F_{c} = F_{B} \rightarrow \frac{m*v^{2}}{r} = qvB$ (1)

<u>Where:</u>

<em>m: is the proton's mass = 1.67*10⁻²⁷ kg</em>

<em>v: is the proton's velocity</em>

<em>r: is the radius of the proton's orbit</em>

<em>q: is the proton charge = 1.6*10⁻¹⁹ C</em>

<em>B: is the magnetic field = 0.040 T </em>

Solving equation (1) for r, we have:

$r = \frac{mv}{qB}$ (2)

By conservation of energy, we can find the velocity of the proton:

$K = U \rightarrow \frac{1}{2}mv^{2} = q*\Delta V$ (3)

<u>Where:</u>

<em>K: is kinetic energy</em>

<em>U: is electrostatic potential energy</em>

<em>ΔV: is the potential difference = 1.0 kV </em>

Solving equation (3) for v, we have:

$v = \sqrt{\frac{2q\Dela V}{m}} = \sqrt{\frac{2*1.6 \cdot 10^{-19} C*1.0 \cdot 10^{3} V}{1.67 \cdot 10^{-27} kg}} = 4.38 \cdot 10^{5} m/s$

Now, by introducing v into equation (2), we can find the radius of the proton's resulting orbit:

$r = \frac{mv}{qB} = \frac{1.67 \cdot 10^{-27} kg*4.38 \cdot 10^{5} m/s}{1.6 \cdot 10^{-19} C*0.040 T} = 0.11 m$

Therefore, the radius of the proton's resulting orbit is 0.11 m.

I hope it helps you!

5 0

3 years ago

What is the only function of the pulleys in the diagram?

tia_tia [17]

The only function of the pulleys in the diagram is to change the direction

of the force applied to raise the bricks.

<h3>What is a Pulley?</h3>

A pulley is a wheel which has a flexible rope on its rim and helps to

transmit energy and motion.

In the diagram given, we can see that the pulley is used to raise a mass of

block by three people. They pull the rope horizontally in order to raise the

block vertically. This means that it was used to change the direction of the

applied force.

Read more about Pulley here brainly.com/question/177456

5 0

3 years ago